I’m sorru but there’s nothing there
For this case what we must do is find a quadratic function that is already factored.
This is because in the factored quadratic equations, it is easier to observe the zeros of the function.
In this case, the zeros of the function represent the time at which the company did not make any profit.
We have the following equation:
p (t) = 40 (t - 3) (t + 2) (t - 5) (t + 3)
We observed that there was no gain in:
t = 3
t = 5
The other roots are discarded because they are negative
Answer:
a.p (t) = 40 (t - 3) (t + 2) (t - 5) (t + 3)
Well - I mean. If every lunch costs $4 that means that the relationship should be linear.
You used to be able to draw graphs. But plot it out.
Round ur quotient and the divsor like 100 divide by 9... would be about 11 and 9 is about 10 and 11 is about 10 so 10 times 10 is one hundred ao the quotient is reasonable
Answer:
b
Step-by-step explanation: